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## Hot answers tagged pde

2

The "Maximum Principle" is a very useful tool to answer to such a question (as John Barber rightly did). I will not come back with this principle to repeat what was already said and which is more general for all times from $t=0$ to $t\to\infty$. In the present case, since the wording of the question concerns only $t\to\infty$ and where it is not question of ...

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Since $\|f - f_n \|_p \to 0$, we can extract a subsequence $f_{n_j}$ so that $\| f - f_{n_j} \|_p \le \frac{1}{2^j}$. Put $$g = \lvert f \rvert + \sum^\infty_{j=1} \lvert f - f_{n_j} \rvert.$$ Then $g \in L^p$ and by the triangle inequality we have $$\lvert f_{n_j} \rvert \le g \,\,\, (\text{almost everywhere}).$$ From this subsequence, you can extract a ...

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Assume that $v_1^2+v_2^2\neq 0$. Write $u(x,y)=U(z,w)$ with $$z:=v_1x+v_2y\text{ and }w:=v_2x-v_1y\,.$$ Hence, $$x=\frac{v_1z+v_2w}{v_1^2+v_2^2}\text{ and }y=\frac{v_2z-v_1w}{v_1^2+v_2^2}\,.$$ Observe that $z$ and $w$ are independent (by checking that $\frac{\partial z}{\partial w}=0$ and $\frac{\partial w}{\partial z}=0$). Then, you can see that $$\frac{\... 1 To be clear, you want to find P_n(t) for a birth-death process with constant birth rate q(i,i+1)=\lambda and constant death rate q(i,i-1)=\mu? This particular birth and death process is exactly the M/M/1 queue. As you can see, under "Transient solution", there is a solution for the probability mass function dependent on time for a particular state. 1 Differentiate both sides of the equation f(\lambda x) = \lambda^{a}f(x) in \lambda to obtain: $$\sum_{i = 1}^{n}x_{i}(\partial_{i}f)(\lambda x) = a\lambda^{a-1}f(x)$$ Setting \lambda = 1 yields \sum_{i = 1}^{n}x_{i}(\partial_{i}f)(x) = af(x), known as Euler's identity. Differentiating the equation displayed above in \... 1 You can find the general statement and proof in Chapter 6 of the book "Sobolev Spaces" by Robert A. Adams and John. J. F. Fournier. 1 Use the Co-area formula with g=|\nabla u|^{-1} \chi_{\Omega_t} to get$$ \int_{\Omega_t} dA=\int_t^\infty \left(\int_{\partial \Omega_x} \frac{ds}{|\nabla u|}\right)dx.$$Differentiating this with respect to t gives the formula you're looking for. 1 Let L be a k^{th}-order linear differential operator, i.e. one which satisfies L(\alpha u + \beta v) = \alpha L u + \beta Lv for all u,v\in C^k and constants \alpha,\beta (of course this notion can be weakened past C^k, but this will do for here). We say that the equation L u =f, for some given function f, is linear. A semilinear equation ... 1 If we set$$x(\xi, t) = f(\xi + \beta t, t) = f(z, t)we find (by the chain rule) \begin{align} x_{t} &= f_{z} \cdot z_{t} + f_{t} \\ &= \beta f_{z} + f_{t} \\ x_{\xi} &= f_{z} \cdot z_{\xi} \\ &= f_{z} \\ x_{\xi \xi} &= f_{zz} \cdot z_{\xi} \\ &= f_{zz} \end{align} Substituting into our original PDE, we get \begin{align} x_{t}... 1 The only way to have a spatially smooth solution is for \partial u/\partial r go to zero as r goes to zero. This means you can use L'Hopitals's rule in evaluating that second term: \displaystyle \frac{1}{r}\frac{\partial u}{\partial r} \rightarrow \frac{\partial_r(\partial u/\partial r)}{\partial_r(r)} = \frac{\partial^2 u}{\partial r^2} 1 You can transform to Sturm-Liouville form by dividing by x: (xM')'+\frac{\lambda}{x}M = 0. $$Whenever you have such a form, you can get rid of the multiplier in the derivative. There is a standard trick when you change the independent variable. In this case, let M(x) = P(\int\frac{1}{x}dx). Then$$ xM'(x) = ...

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$$x^2\frac{d^2M}{dx^2}+x\frac{dM}{dx}+\lambda M(x)=0$$ The change of variable $\quad z=e^x\quad\to\quad x=\ln|z[\quad\to\quad \frac{dz}{dx}=z \quad$ isn't a good idea because it leads to a more complicated form of ODE : $\frac{dM}{dx}=\frac{dM}{dz}\frac{dz}{dx}=z\frac{dM}{dz}$ $\frac{d^2M}{dx^2}=\left(\frac{d}{dz}\frac{dM}{dx}\right)\frac{dz}{dx}=\frac{d}{... 1 Another particular solution you missed can be found as:$\frac{1}{(D+D')^2}x=\frac{1}{D'^2}[1+\frac{D}{D'}]^{-2}x=\frac{1}{D'^2}(1-2D/D'+....]x=\frac{1}{D'^2}(x-2y)=\frac{xy^2}{2}-\frac{y^3}{3}\$

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